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nasty-shy [4]
3 years ago
7

Round to the underlined place.​

Mathematics
1 answer:
nignag [31]3 years ago
8 0

Answer:

158

Step-by-step explanation:

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Use the distributive property to find an expression equivalent to <br> 2x(x+7-(3x+1)
STatiana [176]

Answer:

-4x² + 12x

Step-by-step explanation:

2x(x+7-1(3x+1)) = 2x(x+7-3x-1)

= 2x² + 14x - 6x² - 2x

= -4x² + 12x

8 0
3 years ago
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16-2t=t+9+4t. need help with this​
kkurt [141]

Answer:

Isolate the variable by dividing each side by factors that don't contain the variable.

t =1

3 0
3 years ago
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Show how to divide 9 pieces of licorice among 4 people
Ivanshal [37]

Answer:

1st question,<u> are you are able to cut the four pieces of licorice?</u> <u>if so cut them into 4 equal pieces and divide them among the 4 people</u>, making sure that each person gets an equal 9 pieces.

Step-by-step explanation:

9 pieces of licorice cut equally into 4 smaller bits then distributed throughout 4 people would play as this.

9x4=36

36/4 = 9

Each person would get 9 smaller pieces of licorice

6 0
4 years ago
A rectangular garden is fenced on all sides with 258 feet of fencing. The garden is 7 feet
Bumek [7]

Answer:

L = w + 8

 

P = 2L + 2w

 

256 = 2 (ancho + 8) + 2w

 

256 = 2 vías + 16 + 2 vías

 

256 = 4 semanas + 16

 

240 = 4w

 

w = 60

 

La longitud es 60 + 8 = 68

el ancho es 60

8 0
3 years ago
a bag contains eleven counters. five are white. a counter is taken out the bag and is not replaced. a second counter is taken ou
Fed [463]

Answer:

\displaystyle P(A)=\frac{6}{11}

Step-by-step explanation:

<u>Probabilities</u>

The question describes an event where two counters are taken out of a bag that originally contains 11 counters, 5 of which are white.

Let's call W to the event of picking a white counter in any of the two extractions, and N when the counter is not white. The sample space of the random experience is

\Omega=\{WW,WN,NW,NN\}

We are required to compute the probability that only one of the counters is white. It means that the favorable options are

A=\{WN,NW\}

Let's calculate both probabilities separately. At first, there are 11 counters, and 5 of them are white. Thus the probability of picking a white counter is

\displaystyle \frac{5}{11}

Once a white counter is out, there are only 4 of them and 10 counters in total. The probability to pick a non-white counter is now

\displaystyle \frac{6}{10}

Thus the option WN has the probability

\displaystyle P(WN)=\frac{5}{11}\cdot \frac{6}{10}=\frac{30}{110}=\frac{3}{11}

Now for the second option NW. The initial probability to pick a non-white counter is

\displaystyle \frac{6}{11}

The probability to pick a white counter is

\displaystyle \frac{5}{10}

Thus the option NW has the probability

\displaystyle P(NW)=\frac{6}{11}\cdot \frac{5}{10}=\frac{30}{110}=\frac{3}{11}

The total probability of event A is the sum of both

\displaystyle P(A)=\frac{3}{11}+\frac{3}{11}=\frac{6}{11}

\boxed{\displaystyle P(A)=\frac{6}{11}}

7 0
3 years ago
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